Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 1"
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From <math>4c = \frac{d}4</math> we have that 16 [[divisor | divides]] <math>d</math>. From <math>a + 4 = \frac d4</math> we have <math>d \geq 20</math>. Minimizing <math>d</math> minimizes <math>a, b</math> and <math>c</math> and consequently minimizes our dragon. The smallest possible choice is <math>d = 32</math>, from which <math>a = 4, b = 12</math> and <math>c = 2</math> so our desired number is <math>a + b + c + d = 4 + 12 + 2 + 32 = 050</math>. | From <math>4c = \frac{d}4</math> we have that 16 [[divisor | divides]] <math>d</math>. From <math>a + 4 = \frac d4</math> we have <math>d \geq 20</math>. Minimizing <math>d</math> minimizes <math>a, b</math> and <math>c</math> and consequently minimizes our dragon. The smallest possible choice is <math>d = 32</math>, from which <math>a = 4, b = 12</math> and <math>c = 2</math> so our desired number is <math>a + b + c + d = 4 + 12 + 2 + 32 = 050</math>. | ||
− | + | ==See Also== | |
− | + | {{Mock AIME box|year=2006-2007|n=2|before=First Question|num-a=2}} | |
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[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 09:49, 4 April 2012
Problem
A positive integer is called a dragon if it can be written as the sum of four positive integers and such that Find the smallest dragon.
Solution
From we have that 16 divides . From we have . Minimizing minimizes and and consequently minimizes our dragon. The smallest possible choice is , from which and so our desired number is .
See Also
Mock AIME 2 2006-2007 (Problems, Source) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |